Pak Ho
Lv 5
Pak Ho 發問於 科學及數學數學 · 6 年 前

F.5 Maths Module 2 question

Consider the curve C: x^(2) + 4y^(2) = 68. Find the equation of the normals to C that pass through (0,3)

Thank you :)

3 個解答

評分
  • 6 年 前
    最佳解答

    x^2 + 4y^2 = 68

    Differentiate both sides w.r.t. x, we get

    2x + 8y dy/dx = 0

    ==> dy/dx = -x/(4y)

    So, the slope of the normal at any point on this curve is 4y/x.

    Suppose (x, y) is the required point on this curve, then

    (y - 3)/(x - 0) = 4y/x

    ==> x(y - 3) = 4xy

    ==> x(y - 3) - 4xy = 0

    ==> x(y - 3 - 4y) = 0

    ==> -3x(y + 1) = 0

    ==> x = 0 or y = -1

    When x = 0,

    4y^2 = 68

    ==> y = √17 or y = -√17When y = -1,x^2 + 4(-1)^2 = 68==> x = 8 or -8equation of normal passes thru' (0, 3) and (0, √17) is x = 0;equation of normal passes thru' (0, 3) and (0, -√17) is x = 0;equation of normal passes thru' (0, 3) and (8, -1) is :(y - 3)/(x - 0) = 4*(-1)/8==> x + 2y - 6 = 0equation of normal passes thru' (0, 3) and (-8, -1) is :(y - 3)/(x - 0) = 4*(-1)/(-8)==> x - 2y + 6 = 0

    Therefore, the equation of normals are :(i) x = 0(ii) x + 2y - 6 = 0(iii) x - 2y + 6 = 0

    (Totally there are 3 normals pass thru' (0, 3))

  • 6 年 前

    x^2+〖4y〗^2=68

    2x+4(2y)dy/dx=0

    dy/dx=(-2x)/8y

    dy/dx=(-x)/4y

    When x=0, y=3

    dy/dx=(-0)/3=0

    Equation of the normal to C:

    y-3=(-1)/0(x-0)

    y=3

  • CK
    Lv 7
    6 年 前

    Refer to : http://www.mathportal.org/formulas/analytic-geomet...

    to get the equation of tangent.

    The ellipse is :

    x²/(√68)² + y²/(√17)² = 1

    Tangent at (0,3) is 0x/(√68)² + 3y/(√17)² = 1

    3y = 17

    slope of tangent = 0

    slope of normal is -∞

    equation of normal at (0,3) is x=0;

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